# A charge of #18 C# passes through a circuit every #24 s#. If the circuit can generate #18 W# of power, what is the circuit's resistance?

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The circuit's resistance is approximately (12 , \Omega).

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To find the circuit's resistance, we can use the formula relating power ((P)), current ((I)), and resistance ((R)):

[ P = I^2 \times R ]

First, we need to find the current ((I)). We know that charge ((Q)) passing through a circuit in a given time interval ((t)) can be used to find the current:

[ I = \frac{Q}{t} ]

Given that (Q = 18 , \text{C}) and (t = 24 , \text{s}), we can calculate the current:

[ I = \frac{18 , \text{C}}{24 , \text{s}} = 0.75 , \text{A} ]

Now, we have the current ((I)). We can use this along with the given power ((P = 18 , \text{W})) to find the resistance ((R)) using the formula for power:

[ P = I^2 \times R ]

Plugging in the known values:

[ 18 , \text{W} = (0.75 , \text{A})^2 \times R ]

Solving for (R):

[ R = \frac{18 , \text{W}}{(0.75 , \text{A})^2} ]

[ R = \frac{18 , \text{W}}{0.5625 , \text{A}^2} ]

[ R = 32 , \Omega ]

Therefore, the circuit's resistance is (32 , \Omega).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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